if `A=[[cos,sinx,0],[-sinx,cosx,0],[0,0,1]]=f(x)` then `A^-1`

If A=[{:(cosx,sinx,0),(-sinx,cosx,0),(0,0,1):}] =f(x), then A^(-1) is equal to

If F(x)=[[cosx,-sinx,0],[sinx,cosx,0],[0,0,1]] Show that F(x)F(y)=F(x+y)

If A=f(x)=[(cosx, sinx, 0),(-sinx, cosx,0),(0,0,1)], then the value of A^-1= (A) f(x) (B) -f(x) (C) f(-x) (D) -f(-x)

F(x)=[[cosx,-sinx,0],[sinx,cosx,0],[0,0,1]] and G(x)=[[cosx,0,sinx],[0,1,0],[-sinx,0,cosx]], then [F(x)G(y)]^(-1) is equal to (A) F(-x)G(-y) (B) F(x-1)G(y-1) (C) G(-y)F(-x) (D) G(y^(-1))F(x^(-1))

If f(x) = [(cos x , - sinx,0),(sinx,cosx,0),(0,0,1)] then show f(x) . f(y) = f(x+y)

Let f(x)=[(cosx,-sinx,0),(sinx,cosx,0),(0,0,1)], then (A) (f(x))^2=-I (B) f(x+y)=f(x),f(y) (C) f(x)^-1=f(-x) (D) f(x)^-1=f(x)

f(x)=[(cosx,-sinx,0),(sinx,cosx,0),(0,0,1)] Statement 1: f(x) is inverse of f(-x) Statement 2: f(x).f(y) = f(x+y)

If A=[(2,0,0),(0,cos,sinx),(0,-sinx,cosx)] then (AdjA)^-1= (A) 1/2A (B) A (C) 2A (D) 4A

Find the inverse matrix of the matrix A=|{:(cosx,-sinx,0),(sinx,cosx,0),(0,0,1):}|